btrfs: add little-endian optimized key helpers

The CPU and on-disk keys are mapped to two different structures because
of the endianness. There's an intermediate buffer used to do the
conversion, but this is not necessary when CPU and on-disk endianness
match.

Add optimized versions of helpers that take disk_key and use the buffer
directly for CPU keys or drop the intermediate buffer and conversion.

This saves a lot of stack space accross many functions and removes about
6K of generated binary code:

   text    data     bss     dec     hex filename
1090439   17468   14912 1122819  112203 pre/btrfs.ko
1084613   17456   14912 1116981  110b35 post/btrfs.ko

Delta: -5826

Reviewed-by: Johannes Thumshirn <johannes.thumshirn@wdc.com>
Reviewed-by: Nikolay Borisov <nborisov@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>

1 files changed, 17 insertions, 0 deletions

diff --git a/fs/btrfs/ctree.c b/fs/btrfs/ctree.c
index 82ab6e5a386d..70e49d8d4f6c 100644
--- a/fs/btrfs/ctree.c
+++ b/fs/btrfs/ctree.c
@@ -1501,6 +1501,22 @@ static int close_blocks(u64 blocknr, u64 other, u32 blocksize)
 	return 0;
 }
 
+#ifdef __LITTLE_ENDIAN
+
+/*
+ * Compare two keys, on little-endian the disk order is same as CPU order and
+ * we can avoid the conversion.
+ */
+static int comp_keys(const struct btrfs_disk_key *disk_key,
+		     const struct btrfs_key *k2)
+{
+	const struct btrfs_key *k1 = (const struct btrfs_key *)disk_key;
+
+	return btrfs_comp_cpu_keys(k1, k2);
+}
+
+#else
+
 /*
  * compare two keys in a memcmp fashion
  */
@@ -1513,6 +1529,7 @@ static int comp_keys(const struct btrfs_disk_key *disk,
 
 	return btrfs_comp_cpu_keys(&k1, k2);
 }
+#endif
 
 /*
  * same as comp_keys only with two btrfs_key's